1. A student is asked to use the photoelectric effect to determine the work func
ID: 1273130 • Letter: 1
Question
1. A student is asked to use the photoelectric effect to determine the work function and cutoff wavelength for an unknown sample. Being very lazy, however, the student turns in a lab report stating only that: "When I used 1.49 x1015 Hertz light, the stopping potential was 3.82 Volts." What answers should the student have found for the work function and cutoff wavelength for the sample?
2. When a metal was exposed to light at a frequency of 4.40x1015 s-1, electrons were emitted with a kinetic energy of 4.50x 10-19 J. What is the maximum number of electrons that could be ejected from this metal by a burst of light (at some other frequency) with a total energy of 3.48x10-7 J?
Explanation / Answer
answer to ques 2
Total input energy form the photon = E= h(frequency)
KE= kinetic energy
W = the energy that the nucleus of the metal atom is holding onto the electron
h = planks constant = 6.626 x 10^-34
E= KE +W solve for W so W= hf - KE = 2.273 x 10^ -18 Joules
that property will not change (assuming the same electron) so you can use the value in the second part.
regardless of the amount of energy the burst contains the frequency must be larger than the frequency of the photon that corresponds to the W of the atom ( say f ').
so the cutoff frequency is (f '):
say W= E ' = hf '
then f ' = 3.43 x 10^15 (smallest photon that will eject an electron from the metal)
so if the burst contains 4.46x10^-7 J = (f ' )(h) (n) where n is the number of photons
then n = 1.9618 x 10^ 11 photons at f '
Each photon can only collide, and thus eject, one electron so the maximum number of electrons is also n.
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